Question

In a hydrogen atom, the binding energy of the electron in the nth state is $E_n$,then the frequency of revolution of the electron in the nth orbit is:

• A. $2E_n/nh$
• B. $2E_{n}n/h$
• C. $E_n/nh$
• D. $E_{n}n/h$

Answer 1

The correct option is A $2E_n/nh$

We know that, Kinetic energy=Binding energy

$\frac{1}{2}m{v}^2=E_n----\left(i\right)mvr=\frac{nh}{2\pi }----\left(ii\right)$

Dividing (i) by (ii),

$\frac{v}{2\pi r}=\frac{2E_n}{nh}$

Therefore, the frequency of revolution of the electron in the nth orbit is$=\frac{2E_n}{nh}$